Does there exist a non-parallelisable manifold with exactly $k$ linearly independent vector fields?

Consider $M$, the product of the Klein bottle $K^2$ with $k-1$-torus $T^{k−1}$. This manifold is nonorientable, hence, nonparallelizable. On the other hand, it is the total space of a circle bundle over $T^k$, since $K^2$ is a circle bundle over the circle. Let $H$ be a (smooth) connection on this bundle. Take $k$ independent vector fields $X_1,...X_k$ on $T^k$ and lift them to $M$ via the connection $H$. Hence, $M$ admits $k$ independent vector fields. It cannot have $k+1$ vector fields since it is not parallelizable. I am sure there are orientable examples as well, but this requires more work.

My advisor gave me an example of a closed orientable four-manifold which admits two linearly independent vector fields but not three.

Let $L$ be a complex line bundle over $T^2 = S^1\times S^1$ such that $\langle c_1(L), [T^2]\rangle = 1$; such a line bundle exists because $c_1 : \operatorname{Vect}^1_{\mathbb{C}}(T^2) \to H^2(T^2; \mathbb{Z}) \cong \mathbb{Z}$ is an isomorphism. Then define $M$ to be the total space of the projectivised bundle $\mathbb{P}(L\oplus\varepsilon_{\mathbb{C}}^1)$. Note that $\pi : M \to T^2$ is a $\mathbb{CP}^1$-bundle. We can lift the two linearly independent vector fields on $T^2$ to $M$ via an Ehresmann connection, so $M$ admits at least two linearly independent vector fields

As $L$ is an open subset of $M$, $M$ contains the image of the zero section of $L$ as a compact submanifold, call it $E$. As $\langle e(L), [T^2]\rangle = \langle c_1(L), [T^2]\rangle = 1$, $E$ has self-intersection $1$; in particular, it's mod $2$ intersection number $E\cdot E$ is $1$. On the other hand, the mod $2$ intersection number is also given by $E\cdot E = \langle \alpha\cup\alpha, [T^2]\rangle$ where $\alpha \in H^2(M; \mathbb{Z}_2)$ is the Poincaré dual of the $\mathbb{Z}_2$-fundamental class of $E$. Using the language of Steenrod squares and Wu classes, one can show that $x\cup x = (w_2 + w_1\cup w_1)\cup x$ for all $x \in H^2(M; \mathbb{Z}_2)$; see my answer here for example. As $M$ is orientable, we see that $$1 = E\cdot E = \langle\alpha\cup\alpha, [T^2]\rangle = \langle w_2\cup\alpha, [T^2]\rangle,$$ so $w_2 \neq 0$ (i.e. $M$ is not spin). As such, $M$ cannot admit three linearly independent vector fields; if it did, then $TM \cong \ell\oplus\varepsilon_{\mathbb{R}}^3$ for some real line bundle $\ell$ and hence $w_2 = 0$.

From this construction we obtain, for any $k \geq 2$, an example of a closed, orientable, non-parallelisable manifold such that the maximum number of linearly independent vector fields it admits is $k$, namely $M\times T^{k-2}$. It admits at least $k$ linearly independent vector fields as it is a $\mathbb{CP}^1$-bundle over $T^k$, and it does not admit any more as it is not spin (a product of manifolds is spin if and only if the individual factors are spin). For $k = 1$, we can take $S^5$.